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## Machine Design MCQ Set 1

1. Which one is having highest density?
a) Snow
b) Teflon
c) Soil
d) Sand

Answer: b [Reason:] Teflon, also known as polytetrafluoroethylene has density 2200 kg/m3. It is a thermoplastic polymer, which is a white solid at room temperature. Density of snow, soil and sand are 110 kg/m3, 2050 kg/m3 and 1515 kg/m3 respectively.

2. What is the density of paper in kg/m3?
a) 910
b) 920
c) 930
d) 940

Answer: c [Reason:] Paper density is used in pulp or fabric industry in order to measure of mass of the product per unit of area for a type of paperboard.

3. Which one is having least density?
a) Ice
b) Pyrex
c) Paraffin
d) Marble

Answer: a [Reason:] Ice is water. Depending on the presence of impurities such as particles of soil or bubbles of air, it can appear transparent or a more or less opaque bluish-white color.

4. What is the density of sand in kg/m3?
a) 1215
b) 1315
c) 1415
d) 1515

Answer: d [Reason:] Density of sand can vary depending on the grain size and moisture content and how tightly it is compacted.

5. What is the density of snow in kg/m3?
a) 100
b) 110
c) 120
d) 130

Answer: b [Reason:] Snow is precipitation in the form of flakes of crystalline water ice that falls from clouds. It is a granular material, soft, white and fluffy structure.

6. Which one is having least density?
a) Clay
b) Coal
c) Cotton
d) Banana

Answer: c [Reason:] Cotton is a soft, fluffy staple fiber that grows in a boll. The fiber is spun into yarn or thread. Density of cotton is 80 kg/m3 while that of clay, coal and banana are 1460 kg/m3, 1350 kg/m3 and 980 kg/m3.

7. What is the density of limestone in kg/m3?
a) 2320
b) 2430
c) 2530
d) 2630

Answer: a [Reason:] It is also known as calcium hydroxide. It is a colorless crystal which is obtained from calcium oxide.

8. What is the density of granite in kg/m3?
a) 2130
b) 2230
c) 2530
d) 2630

Answer: d [Reason:] It is granular and phaneritic in nature. This mainly consist of feldspar, mica and quartz.

9. Which one is having highest density?
a) Leather
b) Paper
c) Balsa
d) Snow

Answer: a [Reason:] Density of leather is 998 kg/m3 while that of paper, balsa and snow are 930 kg/m3, 140 kg/m3 and 110 kg/m 3.

10. What is the density of oak in kg/m3?
a) 445
b) 545
c) 645
d) 745

Answer: b [Reason:] Oak has spirally arranged leaves, with lobate margins in many species. There are strong molecular forces of attraction between their bonds.

## Machine Design MCQ Set 2

1. Oxygen diffuses through a stagnant layer of air, 1mm thick, ambient temperature 28°C and 1atm total pressure. The partial pressure of oxygen on two sides of layer is P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.
(a) 2.713
(b )1.713
(c) 0.1713
(d) 0.01713

Answer: b [Reason:] This is a case of diffusion of A through non-diffusing B. Molar flux of oxygen, NA= (DAB*P ln⁡[(P-P2)/(P-P1)])/RTl Putting all the required values in the above equation. NA=1.904 mol/m2.s Molar average velocity, U= (NA+NB)/C U=NA/C (NB=0) U=NART/P U=4.702*10-2m/s Now, UA=UC/CA UA=U/yA=0.4702 m/s (yA is mole fraction of A at the end) JA=CA(uA-U) =PA*( uA-U)/RT =1.713mol/m2.s.

2. In order to prepare Lithium nitride, air is passed gently over a reactor. Suppose that Nitrogen diffuse to a length 5cm before getting reacted with Lithium. Nitrogen reacts with lithium quickly so that partial pressure of Nitrogen at that point is 0. Reactor is a cylinder of length 10cm and diameter 2cm. Assume air to be a mixture of Nitrogen and Oxygen only. The temperature is 298K and total pressure 1atm. Diffusivity of N2 in O2 is 2*10-5m2/s. Calculate the Molar flux (mol/m2.s) of N2 at a distance of 2.5cm from the top with respect to an observer moving with twice the mass average velocity a direction towards the liquid surface. (R=8.205*10-5m3atmK-1mol-1). Assume z=0 at the top of the reactor.
(a) 4.242*10-4
(b) 4.242*10-5
(c) 4.242*10-3
(d) 4.242*10-2

Answer: a [Reason:] Case: Diffusion of A through non-diffusing B N2 in air= 79% O2 in air= 21% Molar flux of N2, NA=0.025mol/m2.s Pressure of N2 at the point z=2.5cm= ( DAB*P ln⁡[(P-P5)/(P-P0)])/RTl=(DAB*P ln⁡[(P-P5)/(P-P2.5)])/RTl P2.5=0.542atm Velocity of N2= NART/P2.5=112.78*10-5m/s Molar flux of O2-=0 Velocity of O2=0 Average molar mass=0.542*28+0.458*32=29.832 Mass average velocity= (0.542*28*112.78*10-5)/29.832=57.37*10-5m/s Required molar flux at the distance of z=5, is CA(uA-2V)= NA-2PAV/RT=4.242*10-4mol/m2.s.

3. A binary liquid mixture is separated by distillation process. The more volatile A vaporizes while less volatile B gets transported in opposite direction. Latent heat of vaporization of A and B is 0.5unit and 1unit respectively. What will be the relation between molar flux of A and B
a) NA=2NB
b) 2NA=NB
c) NA=4NB
d) 4NA=NB

Answer: a [Reason:] Latent heat of vaporization of A is provided from condensation of B. NA∆HA=NB∆HB.

4. In an accident a container containing ethanol(s.g 0.789) falls down in a room of dimension 1*1*2(all in m). Ethanol forms a layer of 2mm. Ethanol vaporizes and diffuses through a stagnant film of air of thickness 2.5mm. What will be the time required for the water layer to disappear completely. The diffusivity of ethanol in air is 2.567*10-5 m2/s. the total pressure is 1.013 bar and the pressure of ethanol in bulk air is 0.02244 bar and at the ethanol-air interface is 0.02718bar. The temperature is 25.2°C.
a) 4.73h
b) 14.6h
c) 5.87h
d) 16.2h

Answer: a [Reason:] This is the case of diffusion of through non diffusing B. NA=DP/RTZ*ln⁡[(P-P1)/(P-P2)] NA=(2.567*10-5*1.013)/(0.08317*298.2*2.5*10-3)*ln⁡[(1.013-0.02244)/(1.013-0.02718)] = 2.01*10-6 kmol/m2.s=9.259*10-5kg/m2.s Amount of ethanol = 2*10-3*1=0.002m3=1.578kg Time for evaporation= 1.578/9.259*10-5=17042.87s=4.73h.

5. In the previous question if the floor has micro pores and water penetrates the floor at a constant rate of 0.1kg/m2h, what will be the time required for the water layer to disappear?
a) 3.64h
b) 4.73h
c) 5.94h
d) 2.36h

Answer: a [Reason:] Combined rate of disappearance= 0.1+9.259*10-5*3600=0.433kg/m2h Time for disappearance= 1.578/0.433=3.64h.

6.The molar average velocity of the components in a binary mixture in which they are in equimolar counter diffusion is
a) Equal to mass average velocity
b) Zero
c) Always negative
d) Always positive

7. A cylinder of sulfur is kept in a large volume of stagnant air for sublimation. How much time will be required for its complete sublimation?
a) One fourth the time required by sphere of same mass
b) Half the time required by a cube of same mass
c) One fourth the time required by a cube of same mass
d) Infinite

Answer: d [Reason:] No or negligible mass transfer in stagnant air.

8. For a gas mixture (A+B) equimolar counterdiffusion takes place. Which among the following statements are correct for molar flux?
a) If temperature is doubled, the flux will get halved.
b) If diffusion length is halved, the flux will get halved.
c) If temperature is halved, the flux will get halved.
d) If temperature is doubled and length is halved, the flux will become four times.

Answer: a [Reason:] NA=(D*(P1-P2))/RTL……….(for equimolar counter diffusion).

## Machine Design MCQ Set 3

1. Consider steady-state equimolar counterdiffusion through the converging-diverging tube. Given: r1=1cm; r2=2cm; NA=10-6 kmol/m2.s at the vessel 1 end of the tube. The rate of transport of B at the vessel 2 end of the tube is
a) 2*10-6 kmol/s, towards vessel 1
b) 2.5*-7 kmol/s, towards vessel 1
c) 5*10-6 kmol/s, towards vessel 2
d) 9*10-7 kmol/s, towards vessel 2

2. Two large vessels containing gaseous mixture of A and B at different concentrations but at the same total pressure are connected by a tapered tube of length 15cm and end diameters 1cm and 4 cm. what should be the diameter of a cylinderical tube of the same length that allows the same rate of transport of A?
a) 2.5 cm
b) 2 cm
c) 3cm
d) 3.5 cm

3. At which point of the following locations inthe tapered tube of the previous problem is the magnitude of the flux maximum?
a) At the larger diameter end
b) At the smaller diameter end
c) Midway in the tube
d) Somewhere in between the midway and the larger diameter end

4. Equimolar counterdiffusion of A and B occurs through a spherical film of i.d=ri and thickness δ. The flux of A is NA1 at r=ri. The flux through a flat film of the same thickness is NA2. if NA1 is larger than NA2 by 5%, the film thickness δ is
a) 0.025ri
b) 0.05ri
c) 0.1ri
d) 0.01ri

5. A napthelene hemisphere is placed on a glass plate and a napthelene ball of identical diameter is kept suspended in the stagnant air in two large rooms. The respective shapes of the ball are maintained while they sublime and reduce in size. Which of the following statements are true in this connection?
a) They take the same time for complete sublimation.
b) The sphere disappears first.
c) The hemisphere disappears first.
d) Nothing definite can be said.

6. Sublimation of a long cylinder made of napthelene occurs in a large volume of stagnant air. The time required for its complete sublimation is
a) infinite
b) half the time of sublimation of a sphere of the same mass
c) twice the time of sublimation of a cylinder of the same mass and lengthy/diameter=1
d) half the time of sublimation of a cylinder of the same mass and lengthy/diameter=1

7. A tapered tube of length 8cm and end diameters 2cm and 4cm connects two vessels containing mixtures of A and B. if the tapered tube is replaced with a cylindrical tube of diameter 6cm, what will be the required length of the cylindrical tube to maintain the same flow rate as in the previous case?
a) 8cm
b) 16cm
c) 36cm
d) 48cm

Answer: c [Reason:] – (Area)*(flux) =Constant=W W1 for cylinder= (πr2 DAB(P1-P2))/RTL W2 for tapered tube = (πDAB*r1*r2(P1-P2))/RTL Equate W1 and W2.

8. In a binary gas mixture containing A and B, If NA=-2NB, what will be the expression for NA?
a) 2(D*(P1-P2))/RTL
b) (D*(P1-P2))/RTL
c) DP*(2RTZ)-1*ln[(P-P1)/(P-P2)].
d) 2DP*(RTZ)-1*ln⁡[(P-P1)/(P-P2)].

Answer: a [Reason:] NA= (NA+N)CA/C – DABdCA/dz NA=(NA/2)PA/P – DAB(dPA/dz) /RT NA(1-2PA/P) = – DAB(dPA/dz)/RT Integrate on both sides.

## Machine Design MCQ Set 4

1. Moisture removal from a material/ compound is known as drying.
a) True
b) False

Answer: a [Reason:] Drying is the moisture removal process from materials.

2. Find the representation X.

a) Water
b) Water vapour
c) Ice
d) All the mentioned

Answer: b [Reason:] During drying operation, the moisture content in the solid gets evaporated in the form of water vapour.

3.The solution can be dried by spraying droplets in a cold gas.
a) True
b) False

Answer: b [Reason:] The moisture in the liquid solution can be removed by spraying the solution droplets in warm/ hot gas.

4. On spraying the liquid over gas leads to ____________
a) Humidification
b) Evaporation
c) Heating of gas
d) None of the mentioned

Answer: b [Reason:] When we spray a liquid over gas it results in evaporation.

5. The removal of acetone from acetone-benzene mixture is drying
a) True
b) False

Answer: b [Reason:] Distillation is the only process we can separate to liquid depends on its volatility.

6. The additional operation requires for drying gas and liquid is _________
a) Absorption
c) Humidification
d) De-humidification

Answer: a [Reason:] Drying includes transfer of solute in the form of water so it requires absorption.

7. The given solid will gain or remove moisture till.

a) pA > P
b) pA< P
c) pA= p
d) pA= 0

Answer: c [Reason:] The operation of heat exchange and solute exchange occurs till the equilibrium occurs ( pA=P).

8. The moisture in the solid after attaining equilibrium is known as
a) Bound moisture
b) Unbound moisture
c) Equilibrium moisture
d) Free moisture

Answer: c [Reason:] Once the equilibrium attain the remaining moisture in solid is because of pA < P of solid so moisture is known as equilibrium moisture content.

9.Find the equilibrium moisture content.

a) 0.2
b) 0.42
c) 0.5
d) None of the mentioned

Answer: d [Reason:] At 100% saturation we get equilibrium moisture content so the value is 0.4.

10. The hysteresis of drying is due to humidification and dehumidification.
a) True
b) False

Answer: b [Reason:] Hysteresis is due to absorption and desorption because these are the operation requires during dying.

11. Find the process of drying in the closed vessel given below.

a) Continuous
b) Batch
c) Semi batch
d) None of the mentioned

Answer: b [Reason:] Here the closed vessel is used so the gas is allowed to contact with solid for some time and removed so it is a batch process.

12. The moisture inside the substance is known as __________
a) Bound moisture
b) Unbound moisture
c) Equilibrium moisture
d) Free moisture

Answer: a [Reason:] After the surface drying i.e. removal of unbound moisture then the inside moisture starts to remove such moisture are known as bound moisture.

13. Find the weight of the wet solid if dry solid is 2 kg and the moisture is 0.5 kg.
a) 2 kg
b) 2.5kg
c) 3 kg
d) 3.5 kg

Answer: b [Reason:] Wet solid= dry solid+ moisture = 2+0.5= 2.5 kg.

14. Find the moisture content in dry basis if the weight of dry solid is 5 kg and the moisture is 2 kg.
a) 0.2
b) 0.3
c) 0.4
d) 0.5

Answer: c [Reason:] As it is a dry basis, the moisture content = 2/5 =0.4.

15.Find the moisture content in wet basis is the weight of the dry solid is 3 kg and the weight of the moisture is 2 kg.
a) 0.2
b) 0.3
c) 0.35
d) 0.4

Answer: d [Reason:] Weight of the solid = weight of moisture/ weight of moisture + dry solid = 2/5 = 0.4.

## Machine Design MCQ Set 5

1. Direct dryers are ___________
a) Batch driers
b) Continuous driers
c) Semi-batch driers
d) None of the mentioned

Answer: a [Reason:] In batch drying process the heat is directly allowed over the solids.

2. Cabinet/ Tray driers are direct driers.
a) True
b) False

Answer: a [Reason:] In cabinet the solids are loaded and the hot gas is passed over. So it’s a kind of direct drying operation.

3. For estimating the drier size it is necessary to know _________
a) Time of drying
b) Heat of drying
c) Speed of drying
d) All of the mentioned

Answer: a [Reason:] The design of drier is based on drying curve rate which demonstrates the time rate.

4. For a batch drying, the wet surface should be more compare to dry surface.
a) True
b) False

Answer: b [Reason:] For an effective batch drying operation to take place, the ratio of wet and dry surface should be same.

5. Constant drying conditions includes
a) Temperature constant
b) Humidity constant
c) Velocity constant
d) All of the mentioned

Answer: d [Reason:] The temperature, humidity and velocity should be constant for constant drying condition.

6. Find the missing position.

a) Time
b) Speed
c) Water vapour
d) None of the mentioned

Answer: a [Reason:] The moisture content decreases with time.

7. Calculate the flux (kg/sq.m sec) if
Mass of dry solid = 8 kg; Wet surface area = 4 sq.m; Change in moisture content with time is 0.4 /sec;
a) 0.4
b) 0.6
c) 0.8
d) 0.9

Answer: c [Reason:] Flux = 8/4 * 0.4 = 0.8

8. The other part than the constant rate period in the Batch- Drying curve is falling rate period.
a) True
b) False

Answer: a [Reason:] After the constant rate period, the unsaturated drying surface is known as falling rate period.

9. Find the gas phase mass transfer co-efficient for a unbound moisture to remove if the flux is 5 kg/sq.m sec and the difference in humidity of the liquid and the main stream is 0.5 units.
a) 5
b) 10
c) 20
d) 30

Answer: b [Reason:] Flux= Mass transfer co-efficient * (Ys-Y) Mass transfer co-efficient= 5/0.5 = 10

10. The value of _________ remains constant while drying if speed and direction of gas flow never changes.
a) Mass transfer co-efficient
b) Humidity
c) Moisture
d) None of the mentioned

Answer: a [Reason:] If the speed and the direction of gas never changes the rate remains same so the mass transfer co-efficient remain same.

11. If the dry spot appears in the substance in the batch drying curve at ___________
a) Critical moisture content
b) Equilibrium moisture content
c) Bound moisture
d) Unbound moisture

Answer: a [Reason:] After the constant drying period, the moisture of unbound will finish then the dry spots appears. The constant drying period ends at critical moisture content.

12. After critical moisture content _________ starts.
a) Saturated drying region
b) Unsaturated drying region
c) Constant drying region
d) None of the mentioned

Answer: b [Reason:] After the critical moisture content, the falling period starts which are the unsaturated drying region.

13. After the unsaturated drying completed ________ starts to evaporate.
a) Bound
b) Unbound
c) Equilibrium
d) None of the mentioned

Answer: a [Reason:] After the unsaturated drying the internal movement starts which is the evaporation of bound moisture.

14.Find the representation AB in the batch drying curve.

a) Constant drying region
b) Falling rate period
c) Equilibrium moisture content
d) None of the mentioned